Derivatives: change at a point

A function tells you an output. Its derivative tells you how quickly that output changes when you change the input. For $f(x)=x^2$, these are different questions:

• What is the output at 3? $f(3)=9$.
• How fast is the output changing there? We will find $f'(3)=6$.

Rise over run

Start at $x=3$ and move to $3+h$. The letter $h$ names the input change; $\Delta x$ is another common notation for the same thing. The output changes from $f(3)$ to $f(3+h)$.

Divide output change by input change:

$$\frac{f(3+h)-f(3)}{h}.$$

This is an average rate over an interval, or the slope of the line joining two points. For $h=1$, the output goes from 9 to 16, so the average rate is 7.

Shrink the step

A smaller interval tells us more about the rate near 3:

Input change $h$	Output change $(3+h)^2-9$	Average rate
1	7	7
0.1	0.61	6.1
0.01	0.0601	6.01
−0.01	−0.0599	5.99

The rates approach 6 from either side. Expand the numerator to see why:

$$\frac{(3+h)^2-9}{h} =\frac{6h+h^2}{h} =6+h\qquad(h\ne0).$$

For every nonzero $h$, this is an ordinary division. As $h$ approaches zero, $6+h$ approaches 6. We never substitute zero into the denominator.

A limit records that approaching value. The derivative at 3 is

$$f'(3)=\lim_{h\to0}\frac{f(3+h)-f(3)}h=6.$$

Read $f'(3)$ as “f prime at 3.” The notation $df/dx$ also denotes the derivative; it identifies the output f and the input x being varied.

Zoom in

Near a differentiable point, the curve is well approximated by a straight line. Here that line passes through $(3,9)$ with slope 6.

At x = 3, shorten the step to compare the average rate 6 + h with the derivative 6. Zooming shows the curve approaching its tangent line locally.

The slope is a function

Repeat the calculation at an arbitrary input x:

$$\frac{(x+h)^2-x^2}{h}=2x+h\longrightarrow2x.$$

So $f'(x)=2x$. This is a function of x. Evaluating it at 3 gives the number $f'(3)=6$.

For comparison, a constant function has derivative zero. A straight line $f(x)=3x+1$ has derivative 3 at every point.

What a derivative is for

Multiply the rate by a small input change to estimate the output change:

$$f(x+h)\approx f(x)+f'(x)h.$$

At 3, the predicted output at 3.01 is $9+6(0.01)=9.06$. The exact output is 9.0601. The omitted term is $h^2=0.0001$.

Keep the three quantities separate: 6 is the rate, 0.06 is the predicted change, and 9.06 is the predicted new output. If x measured seconds and f measured metres, their units would be metres per second, metres, and metres respectively.

Where there is no slope

For $f(x)=|x|$ at zero, the average rate is 1 for every positive h and −1 for every negative h. There is no common limit, so the derivative does not exist there. A corner does not acquire a unique tangent just because we zoom in.

Try it: estimate the change to x = 2.99

Use $h=-0.01$. The estimated change is $6(-0.01)=-0.06$, giving an output of 8.94. The exact output is 8.9401. A positive derivative predicts a decrease when the input step is negative.

Next

Derivative rules let us combine known derivatives without repeating the limit calculation every time.